3rd order homogeneous Linear ODE matrix transformation

[pat666]

Homework Statement

Transform this 3rd order homogeneous linear ODE with constant coefficients using matrix notation.
y'''+7y''+6y'+3y=0

Homework Equations

The Attempt at a Solution

I can't find anything useful to start with on this one. I need some with starting this one, I've been reading a wiki article but it doesn't make much sense to me..

Thanks for any help

 

[lanedance]

so i guess you want to transform into a coupled linear system of first order DEs by letting
x_{i+1} = x_i'

then the xi are related by a constant matrix A
\textbf{x}'=A\textbf{x}

 

[Metaleer]

You need to redefine some variables.

First write your differential equation as y''' = -7 y'' - 6y - 3y. Now you define

y_1 = y,

\boxed{y_2} = y' = \boxed{y_1'},

\boxed{y_3} = y'' = \boxed{y_2'},​

and from the ODE,

\boxed{y_3' = - 7y_3 - 6 y_2 - 3y_1.}​

With the last three equations (boxed parts), you can write out a linear system of differential equations, where the primed variables are the nonhomogeneous terms, and the unknowns are the unprimed variables. From here, you just write the set of equations in matrix form.

Hope this helps. :)

 

[pat666]

Hi,
This looks very familiar. Primed variables are y' and y'', unprimed -7,-6 and -3. What do I do with these? I know it involves matrices but how do I set it up?

Thanks

 

[lanedance]

so let
x_1 = y
x_2 = x_1' = y'
x_3= x_2' = y''

then write
\textbf{x} = (x_1,x_2,x_3)^T

then the system of equations can be written as
\textbf{x}'=A \textbf{x}

try and fill in the entries of A yourself

 

[pat666]

So A will be (-7,-6,-3)?
thanks

 

[lurflurf]

y'''+7y''+6y'+3y=(D^3+7D^2+6D+3)y=0
we want to write D as a matrix (D=A), chose the basis {y,y',y''}
a matrix is defined by its action on a basis
Dy''=-7y''-6y'-3y
Dy' = y''
Dy = y'
Which should be easy to write in matrix form
Another approch would be to note
(D^3+7D^2+6D+3)(a y''+b y'+c y)=0
thus A has (D^3+7D^2+6D+3) for a charateristic polynomial and construct A accordingly.

 

[lurflurf]

So A will be (-7,-6,-3)?

The most interesting row will be. A should have 9 entries, 3rows 3 columns.

 

[pat666]

then the rest would be:
(1,0,0)
(0,1,0)
(-7,-6,-3)
*
<br /> \textbf{x} = (x_1,x_2,x_3)^T <br />

Does the ^T mean transpose?

 

[lurflurf]

That is right, and ^ does mean transpose.

 

[pat666]

I get x'=
(x_1)
(x_2)
(-7x_1-6x_2-3x_3)

What do I do now that I have x'

Thanks